a) A = (sqrt(7) + sqrt(3))/(sqrt(7) - sqrt(3)) + (sqrt(7) - sqrt(3))/(sqrt(7) + sqrt(3)) b) B = 2sqrt(27) + sqrt((1 - sqrt(3)) ^ 2) - 4/(sqrt(2))
Bài 1. (2,0 điểm) Thực hiện phép tính: n) 7/9 * sqrt(81) - 1/2 * sqrt(16) . c) (sqrt(8/3) - sqrt(24) + sqrt(50/3)) , sqrt 12 . » sqrt((sqrt(7) - 4) ^ 2) + sqrt(7) 1/(5 + 2sqrt(3)) + 1/(5 - 2sqrt(3))
a) \(\sqrt{4x^2-9}=2\sqrt{x+3}\)
b) \(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
c) \(\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27\sqrt{\dfrac{x-1}{81}}=4\)
d)\(5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
tính
A=\(\left(1-\sqrt{7}\right).\dfrac{\sqrt{7}+7}{2\sqrt{7}}\)
B=\(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
C=\(\sqrt{32}-\sqrt{50}+\sqrt{18}\)
D=\(\sqrt{72}+\sqrt{4\dfrac{1}{2}}-\sqrt{32}-\sqrt{162}\)
E=\(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
a : \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b : \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
c : \(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
d : \(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\)
a.
$A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}$
$A\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}$
$A\sqrt{2}=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}$
$=|\sqrt{3}-1|+|\sqrt{3}+1|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}$
$\Rightarrow A=2\sqrt{3}: \sqrt{2}=\sqrt{6}$
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$B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$
$B\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$
$B\sqrt{2}=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$
$=|\sqrt{7}-1|-|\sqrt{7}+1|=\sqrt{7}-1-(\sqrt{7}+1)=-2$
$\Rightarrow B=-2:\sqrt{2}=-\sqrt{2}$
\(a,\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(A-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)
Ta có :
\(A-\sqrt{2}=-2\\ \Leftrightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)
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C làm giống câu a, nhé.
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\(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}+1\right|-\left|\sqrt{5}-2\right|\\ =2\sqrt{5}+1-\sqrt{5}+2\\ =3+\sqrt{5}\)
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\(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\\ =\sqrt{48-2\cdot4\cdot\sqrt{3}\cdot2+4}+\left|4\sqrt{3}-7\right|\\ =\sqrt{\left(4\sqrt{3}\right)^2-2\cdot4\cdot\sqrt{3}\cdot2+2^2}+4\sqrt{3}-7\\ =\sqrt{\left(4\sqrt{3}-2\right)^2}+4\sqrt{3}-7\\ =4\sqrt{3}-2+4\sqrt{3}-7\\ =8\sqrt{3}-9\)
c.
$C=\sqrt{(2\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|2\sqrt{5}+1|-|\sqrt{5}-2|=2\sqrt{5}+1-(\sqrt{5}-2)=\sqrt{5}+3$
d.
$D=\sqrt{52-16\sqrt{3}}+\sqrt{4\sqrt{3}-7)^2}$
$=\sqrt{(4\sqrt{3})^2-2.4\sqrt{3}.2+2^2}+|4\sqrt{3}-7|$
$=\sqrt{(4\sqrt{3}-2)^2}+|4\sqrt{3}-7|$
$=|4\sqrt{3}-2|+|4\sqrt{3}-7|$
$=4\sqrt{3}-2+7-4\sqrt{3}=5$
a) 2sqrt(25(x - 3)) - 1/2 * sqrt(4x - 12) + 1/7 * sqrt(49(x - 3)) = 20 b) sqrt(x ^ 2 - 6x + 9) = 2
1. A=\(\sqrt{4+\sqrt{ }7}\)+ \(\sqrt{4-\sqrt{ }7}\) 2. B=\(\dfrac{\sqrt{\sqrt{7-\sqrt{3}}}-\sqrt{\sqrt{7+\sqrt{3}}}}{\sqrt{\sqrt{7}-\sqrt{2}}}\) 3. C=\(\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\) 4. D=\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\) 5. so sánh Cho A=\(\sqrt{11+\sqrt{96}}\) B=\(\dfrac{2\sqrt{2}}{\sqrt{1+\sqrt{2-\sqrt{3}}}}\) so sánh A và B
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
a)\(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
b) \(\sqrt{7+4\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
c) \(\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)
d)\(\sqrt{7+2\sqrt{10}}-\sqrt{3-2\sqrt{2}}\)
a) \(=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) \(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}=2+\sqrt{3}-1-\sqrt{3}=1\)
c) \(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1+\sqrt{7}-1=2\sqrt{7}\)
d) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{2}+1=\sqrt{5}+1\)
Tính:
a) \(\sqrt{27}+\sqrt{75}-\sqrt{\dfrac{1}{3}}\)
b) \(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
c) \(\dfrac{3}{\sqrt{7}+\sqrt{2}}+\dfrac{2}{3+\sqrt{7}}+\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)
#Giải:
a)\(\sqrt{27}\)+\(\sqrt{75}\)-\(\sqrt{\dfrac{1}{3}}\)=8\(\sqrt{3}\)-\(\sqrt{\dfrac{1}{3}}\)=\(\dfrac{23\sqrt{3}}{3}\).
b)\(\sqrt{4+2\sqrt{3}}\)-\(\sqrt{4-2\sqrt{3}}\)=2.
c)\(\dfrac{3}{\sqrt{7}+\sqrt{2}}\)+\(\dfrac{2}{3+\sqrt{7}}\)+\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)=1,093+\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)=2,507.
a) = \(3\sqrt{3}+5\sqrt{3}-\dfrac{1}{\sqrt{3}}\)
= \(3\sqrt{3}+5\sqrt{3}-\dfrac{3}{\sqrt{3}}\)
= \(\dfrac{23\sqrt{3}}{3}\)
b) = \(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
= \(1+\sqrt{3}-\left(\sqrt{3}-1\right)\)
= \(1+\sqrt{3}-\sqrt{3}+1\)
= 2
c) = \(\dfrac{3\left(\sqrt{7}-\sqrt{2}\right)}{5}+\dfrac{2\left(3-\sqrt{7}\right)}{2}+\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)\)
= \(3\sqrt{7}-3\sqrt{2}+3-\sqrt{7}+2\sqrt{2}+2-2-\sqrt{2}\)
= \(\dfrac{3\sqrt{7}-3\sqrt{2}}{5}+3-\sqrt{7}+\sqrt{2}\)
= \(\dfrac{3\sqrt{7}-3\sqrt{2}-5\sqrt{7}+5\sqrt{2}}{5}+3\)
= \(\dfrac{-2\sqrt{7}+2\sqrt{2}}{5}+3\)
\(\approx2,5\)
Tính :
a) \(\dfrac{7\sqrt{3}-\:3\sqrt{7}}{\sqrt{7}-\sqrt{3}}+\dfrac{4}{5-\sqrt{21}}-\:\dfrac{6\sqrt{7}}{\sqrt{3}}\)
b) \(\sqrt{\dfrac{2+2\sqrt{3}}{\sqrt{3}-\:1}}+\dfrac{\sqrt{2}}{1+\sqrt{3}}\sqrt{2\:-\sqrt{3}}\)
c) \(\sqrt{2\:-\sqrt{3}}.\:\left(\sqrt{2+\sqrt{3}}+\sqrt{2}\right)\)
a)\(\dfrac{\sqrt{21}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{7}-\sqrt{3}}+\dfrac{4\left(5+\sqrt{21}\right)}{4}-\dfrac{\sqrt{3}.\sqrt{2}.\sqrt{7}}{\sqrt{3}}\)=\(5+2\sqrt{21}-\sqrt{14}\)
c) (\(\sqrt{2-\sqrt{3}}.\sqrt{2+\sqrt{3}}\))+\(\sqrt{2}\left(\sqrt{2-\sqrt{3}}\right)\)=1+\(\sqrt{2\sqrt{2}-\sqrt{6}}\)